Advanced Trigonometric Proof Question

[agile gazelle]

Not bothered spending 20 minutes scratching my head over this. Anyone know how to solve it?

[vestal flame]

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[naive mango]

Combine them into the same fraction

And write everything in terms of sin and cos

@agile gazelle

[agile gazelle]

whopper whopper whopper whopper

alr

[naive mango]

$\frac{\sec^2{x}-1+\tan^2{x}}{\tan{x}(\sec{x}+1)}$

[tropic lava]

Ephesians 2:8-9

[agile gazelle]

did that already

[naive mango]

You should get that

Okay now everything in terms of sin and cos

[agile gazelle]

yep

hmmmmmmmmmmmmmmmmmmmmm

forgor how

¯_(ツ)_/¯

you mean this?

[naive mango]

$\frac{\frac{1}{\cos^2x}-1+\frac{\sin^2x}{\cos^2x}}{\frac{\sin{x}}{\cos{x}}}(\frac{1}{\cos{x}}+1)}$

Almost

[agile gazelle]

ohhhh that

alr

[tropic lava]

Ephesians 2:8-9
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[naive mango]

Lol

This is so hard on phone

[agile gazelle]

yees

[naive mango]

There, you should get that

Everythin in terms of sin and cos

sec(x)=1/cos(x)

And tan(x)=sin(x)/cos(x)

And just add the square where it's squared

Next we're gonna clean up the numerator so it's 1 fraction

[agile gazelle]

I got this thing

[naive mango]

Like that yeah?

[agile gazelle]

no I simplified it

oops I did something wrong

gimme a sec

[naive mango]

Okay

[agile gazelle]

alr done

[naive mango]

On top, turn 1 into sin^2x +cos^2x

And on the bottom factor out sinx

Top will give you 2sin^2(x)

Bottom sinx(1+cosx)

Cancel out the square and sin and that's it

[agile gazelle]

alr noice

thanks

[naive mango]

Anytime

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